ICSE Solutions for Selina Concise Chapter 5 Quadratic Equation Class 10 Maths
Exercise 5(A)
1. Find which of the following equations are quadratic:
(i) (3x – 1)2 = 5(x + 8)
(ii) 5x2 – 8x = -3(7 – 2x)
(iii) (x -4) (3x + 1) = (3x – 1) (x + 2)
(iv) x2 + 5x – 5 = (x – 3)2
(v) 7x3 – 2x2 + 10 = (2x – 5)2
(vi) (x – 1)2 + (x + 2)2 + 3(x + 1) = 0
Solution
(i) (3x – 1)2 = 5(x + 8)
⇒ (9x2 – 6x + 1) = 5x + 40
⇒ 9x2 – 11x – 39 = 0; which is of the general form ax2 + bx + c = 0.
Thus, the given equation is a quadratic equation.
(ii) 5x2 – 8x = -3(7 – 2x)
⇒ 5x2 – 8x = 6x – 21
⇒ 5x2 – 14x + 21 = 0; which is of the general form ax2 + bx + c = 0.
Thus, the given equation is a quadratic equation.
(iii) (x – 4) (3x + 1) = (3x – 1) (x +2)
⇒ 3x2 + x – 12x – 4 = 3x2 + 6x – x – 2
⇒ 16x + 2 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a linear equation.
Thus, the given equation is not a quadratic equation.
(iv) x2 + 5x – 5 = (x – 3)2
⇒ x2 + 5x – 5 = x2 – 6x + 9
⇒ 11x – 14 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a linear equation.
Thus, the given equation is not a quadratic equation.
(v) 7x3 – 2x2 + 10 = (2x – 5)2
⇒ 7x3 – 2x2 + 10 = 4x2 – 20x + 25
⇒ 7x3 – 6x2 + 20x – 15 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a cubic equation.
Thus, the given equation is not a quadratic equation.
(vi) (x – 1)2 + (x + 2)2 + 3(x +1) = 0
⇒ x2 – 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0
⇒ 2x2 + 5x + 8 = 0; which is of the general form ax2 + bx + c = 0.
Thus, the given equation is a quadratic equation.
2. (i) Is x = 5 a solution of the quadratic equation x2 – 2x – 15 = 0?
Solution
Given quadratic equation, x2 – 2x – 15 = 0
We know that, for x = 5 to be a solution of the given quadratic equation it should satisfy the equation.
Now, on substituting x = 5 in the given equation, we have
L.H.S = (5)2 – 2(5) – 15
= 25 – 10 – 15
= 0
= R.H.S
∴ x = 5 is a solution of the given quadratic equation x2 – 2x – 15 = 0
(ii) Is x = -3 a solution of the quadratic equation 2x2 – 7x + 9 = 0?
Solution
Given quadratic equation, 2x2 – 7x + 9 = 0
We know that, for x = -3 to be solution of the given quadratic equation it should satisfy the equation.
Now, on substituting x = 5 in the given equation, we have
L.H.S = 2(-3)2 – 7(-3) + 9
= 18 + 21 + 9
= 48
≠ R.H.S
∴ x = -3 is not a solution of the given quadratic equation 2x2 – 7x + 9 = 0.
Exercise 5(B)
1. Without solving, comment upon the nature of roots of each of the following equations:
(i) 7x2 – 9x +2 = 0
(ii) 6x2 – 13x +4 = 0
(iii) 25x2 – 10x + 1= 0
(iv) x2 + 2√3x – 9 = 0
(v) x2 – ax – b2 = 0
(vi) 2x2 + 8x + 9 = 0
Solution
(i) Given quadratic equation, 7x2 – 9x + 2 = 0
Here, a = 7, b = -9 and c = 2
So, the Discriminant (D) = b2 – 4ac
D = (-9)2 – 4(7)(2)
= 81 – 56
= 25
As D > 0, the roots of the equation is real and unequal.
(ii) Given quadratic equation, 6x2 – 13x + 4 = 0
Here, a = 6, b = -13 and c = 4
So, the Discriminant (D) = b2 – 4ac
D = (-13)2 – 4(6)(4)
= 169 – 96
= 73
As D > 0, the roots of the equation is real and unequal.
(iii) Given quadratic equation, 25x2 – 10x + 1 = 0
Here, a = 25, b = -10 and c = 1
So, the Discriminant (D) = b2 – 4ac
D = (-10)2 – 4(25)(1)
= 100 – 100
= 0
As D = 0, the roots of the equation is real and equal.
(iv) Given quadratic equation, x2 + 2√3x – 9 = 0
Here, a = 1, b = 2√3 and c = -9
So, the Discriminant (D) = b2 – 4ac
D = (2√3)2 – 4(1)(-9)
= 12 + 36
= 48
As D > 0, the roots of the equation is real and unequal.
(v) Given quadratic equation, x2 – ax – b2 = 0
Here, a = 1, b = -a and c = -b2
So, the Discriminant (D) = b2 – 4ac
D = (a)2 – 4(1)(-b2)
= a2 + 4b2
a2 + 4b2 is always positive value.
Thus D > 0, and the roots of the equation is real and unequal
(vi) Given quadratic equation, 2x2 + 8x + 9 = 0
Here, a = 2, b = 8 and c = 9
So, the Discriminant (D) = b2 – 4ac
D = (8)2 – 4(2)(9)
= 64 – 72
= -8
As D < 0, the equation has no roots.
2. Find the value of ‘p’, if the following quadratic equations has equal roots:
(i) 4x2 – (p – 2)x + 1 = 0
(ii) x2 + (p – 3)x + p = 0
Solution
(i) 4x2 – (p – 2)x + 1 = 0
Here, a = 4, b = -(p – 2), c = 1
Given that the roots are equal,
So, Discriminant = 0 ⇒ b2– 4ac = 0
D = (-(p – 2))2 – 4(4)(1) = 0
⇒ p2 + 4 – 4p – 16 = 0
⇒ p2 – 4p – 12 = 0
⇒ p2 – 6p + 2p – 12 = 0
⇒ p(p – 6) + 2(p – 6) = 0
⇒ (p + 2)(p – 6) = 0
⇒ p + 2 = 0 or p – 6 = 0
Hence, p = -2 or p = 6
(ii) x2 + (p – 3)x + p = 0
Here, a = 1, b = (p – 3), c = p
Given that the roots are equal,
So, Discriminant = 0 ⇒ b2– 4ac = 0
D = (p – 3)2 – 4(1)(p) = 0
⇒ p2 + 9 – 6p – 4p = 0
⇒ p2– 10p + 9 = 0
⇒ p2-9p – p + 9 = 0
⇒ p(p – 9) – 1(p – 9) = 0
⇒ (p -9)(p – 1) = 0
⇒ p – 9 = 0 or p – 1 = 0
Hence, p = 9 or p = 1
Exercise 5(C)
Solve equations, number 1 to 20, given below, using factorization method:
1. x2 – 10x – 24 = 0
Solution
Given equation, x2 – 10x – 24 = 0
⇒ x2 – 12x + 2x – 24 = 0
⇒ x(x – 12) + 2(x – 12) = 0
⇒ (x + 2)(x – 12) = 0
So, x + 2 = 0 or x – 12 = 0
Hence, x = -2 or x = 12
2. x2 – 16 = 0
Solution
Given equation, x2 – 16 = 0
⇒ x2 + 4x – 4x + 16 = 0
⇒ x(x + 4) -4(x + 4) = 0
⇒ (x – 4) (x + 4) = 0
So, (x – 4) = 0 or (x + 4) = 0
Hence, x = 4 or x = -4
3. 2x2 – ½ x = 0
Solution
Given equation, 2x2 – ½ x = 0
⇒ 4x2 – x = 0
⇒ x(4x – 1) = 0
So, either x = 0 or 4x – 1 = 0
Hence, x = 0 or x = ¼
4. x(x – 5) = 24
Solution
Given equation, x(x – 5) = 24
⇒ x2 – 5x = 24
⇒ x2 – 5x – 24 = 0
⇒ x2 – 8x + 3x – 24 = 0
⇒ x(x – 8) + 3(x – 8) = 0
⇒ (x + 3)(x – 8) = 0
So, x + 3 = 0 or x – 8 = 0
Hence, x = -3 or x = 8
5. 9/2 x = 5 + x2
Solution
Given equation, 9/2 x = 5 + x2
On multiplying by 2 both sides, we have
9x = 2(5 + x2)
⇒ 9x = 10 + 2x2
⇒ 2x2 – 9x + 10 = 0
⇒ 2x2 – 4x – 5x + 10 = 0
⇒ 2x(x – 2) – 5(x – 2) = 0
⇒ (2x – 5)(x -2) = 0
So, 2x – 5 = 0 or x – 2 = 0
Hence, x = 5/2 or x = 2
6. 6/x = 1 + x
Solution
Given equation, 6/x = 1 + x
On multiplying by x both sides, we have
6 = x(1 + x)
⇒ 6 = x + x2
⇒ x2 + x – 6 = 0
⇒ x2 + 3x – 2x – 6 = 0
⇒ x(x + 3) – 2(x + 3) = 0
⇒ (x – 2) (x + 3) = 0
So, x – 2 = 0 or x + 3 = 0
Hence, x = 2 or x = -3
7. x = (3x + 1)/ 4x
Solution
Given equation, x = (3x + 1)/ 4x
On multiplying by 4x both sides, we have
4x(x) = 3x + 1
⇒ 4x2 = 3x + 1
⇒ 4x2 – 3x – 1 = 0
⇒ 4x2 – 4x + x – 1 = 0
⇒ 4x(x – 1) + 1(x – 1) = 0
⇒ (4x + 1) (x – 1) = 0
So, 4x + 1 = 0 or x – 1 = 0
Hence, x = -1/4 or x = 1
8. x + 1/x = 2.5
Solution
Given equation, x + 1/x = 2.5
x + 1/x = 5/2
Taking LCM on L.H.S, we have
(x2 + 1)/ x = 5/2
⇒ 2(x2 + 1) = 5x
⇒ 2x2 + 2 = 5x
⇒ 2x2 – 5x + 2 = 0
⇒ 2x2 – 4x – x + 2 = 0
⇒ 2x(x – 2) -1(x – 2) = 0
⇒ (2x – 1)(x – 2) = 0
So, 2x – 1 = 0 or x – 2 = 0
Hence, x = ½ or x = 2
9. (2x – 3)2 = 49
Solution
Given equation, (2x – 3)2 = 49
Expanding the L.H.S, we have
4x2 – 12x + 9 = 49
⇒ 4x2 – 12x – 40 = 0
Dividing by 4 on both side
x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0
⇒ (x + 2) (x – 5) = 0
So, x + 2 = 0 or x – 5 = 0
Hence, x = -2 or 5
10. 2(x2 – 6) = 3(x – 4)
Solution
Given equation, 2(x2 – 6) = 3(x – 4)
2x2 – 12 = 3x – 12
⇒ 2x2 = 3x
⇒ x(2x – 3) = 0
So, x = 0 or (2x – 3) = 0
Hence, x = 0 or x = 3/2
11. (x + 1) (2x + 8) = (x + 7) (x + 3)
Solution
Given equation, (x + 1) (2x + 8) = (x + 7) (x + 3)
2x2 + 2x + 8x + 8 = x2 + 7x + 3x + 21
⇒ 2x2 + 10x + 8 = x2 + 10x + 21
⇒ x2 = 21 – 8
⇒ x2 – 13 = 0
⇒ (x – √13) (x + √13) = 0
So, x – √13 = 0 or x + √13 = 0
Hence, x = – √13 or x = √13
12. x2 – (a + b)x + ab = 0
Solution
Given equation, x2 – (a + b)x + ab = 0
x2 – ax – bx + ab = 0
⇒ x(x – a) – b(x – a) = 0
⇒ (x – b) (x – a) = 0
So, x – b = 0 or x – a = 0
Hence, x = b or x = a
13. (x + 3)2 – 4(x + 3) – 5 = 0
Solution
Given equation, (x + 3)2 – 4(x + 3) – 5 = 0
⇒ (x2 + 9 + 6x) – 4x – 12 – 5 = 0
⇒ x2 + 2x – 8 = 0
⇒ x2 + 4x – 2x – 8 = 0
⇒ x(x + 4) – 2(x – 4) = 0
⇒ (x – 2)(x + 4) = 0
So, x – 2 = 0 or x + 4 = 0
Hence, x = 2 or x = -4
14. 4(2x – 3)2 – (2x – 3) – 14 = 0
Solution
Given equation, 4(2x – 3)2 – (2x – 3) – 14 = 0
Let substitute 2x – 3 = y
Then the equation becomes,
4y2 – y – 14 = 0
⇒ 4y2 – 8y + 7y – 14 = 0
⇒ 4y(y – 2) + 7(y – 2) = 0
⇒ (4y + 7)(y – 2) = 0
So, 4y + 7 = 0 or y – 2 = 0
Hence, y = -7/4 or y = 2
But we have taken y = 2x – 3
Thus,
2x – 3 = -7/4 or 2x – 3 = 2
⇒ 2x = 5/ 4 or 2x = 5
⇒ x = 5/8 or x = 5/2
15. 3x – 2/ 2x- 3 = 3x – 8/ x + 4
Solution
Given equation, 3x – 2/ 2x- 3 = 3x – 8/ x + 4
On cross-multiplying we have,
(3x – 2)(x + 4) = (3x – 8)(2x – 3)
⇒ 3x2 – 2x + 12x – 8 = 6x2 – 16x – 9x + 24
⇒ 3x2 + 10x – 8 = 6x2 – 25x + 24
⇒ 3x2 – 35x + 32 = 0
⇒ 3x2 – 3x – 32x + 32 = 0
⇒ 3x(x – 1) – 32(x – 1) = 0
⇒ (3x – 32)(x – 1) = 0
So, 3x – 32 = 0 or x – 1 = 0
Hence, x = 32/3 or x = 1
16. 2x2 – 9x + 10 = 0, when:
(i) x ∈ N
(ii) x ∈ Q
Solution
Given equation, 2x2 – 9x + 10 = 0
2x2 – 4x – 5x + 10 = 0
⇒ 2x(x – 2) – 5(x – 2) = 0
⇒ (2x – 5)(x – 2) = 0
So, 2x – 5 = 0 or x – 2 = 0
Hence, x = 5/2 or x = 2
(i) When x ∈ N
x = 2 is the solution.
(ii) When x ∈ Q
x = 2, 5/2 are the solutions
17. (x-3)/(x+3) + (x+3)/(x-3) = 2(1/2)
Solution
⇒ 2(2x2 + 18) = 5(x2 – 9)
⇒ 4x2 + 36 = 5x2 – 45
⇒ x2 – 81 = 0
⇒ (x – 9)(x + 9) = 0
So, x – 9 = 0 or x + 9 = 0
Hence, x = 9 or x = -9
Exercise 5(D)
1. Solve, each of the following equations, using the formula:
(i) x2 – 6x = 27
Solution
Given equation, x2 – 6x = 27
x2 – 6x – 27 = 0
Here, a = 1 , b = -6 and c = -27
By quadratic formula, we have
∴ x = 9 or -3
(ii) x2 – 10x + 21 = 0
Solution
Given equation, x2 – 10x + 21 = 0
Here, a = 1, b = -10 and c = 21
By quadratic formula, we have
∴ x = 7 or x = 3
(iii) x2 + 6x – 10 = 0
Solution
Given equation, x2 + 6x – 10 = 0
Here, a = 1, b = 6 and c = -10
By quadratic formula, we have
∴ x = -3 + √19 or x = -3 – √19
(iv) x2 + 2x – 6 = 0
Solution
Given equation, x2 + 2x – 6 = 0
Here, a = 1, b = 2 and c = -6
By quadratic formula, we have
∴ x = -1 + √7 or x = -1 – √7
(v) 3x2 + 2x – 1 = 0
Solution
Given equation, 3x2 + 2x – 1 = 0
Here, a = 3, b = 2 and c = -1
By quadratic formula, we have
∴ x = 1/3 or x = -1
(vi) 2x2 + 7x + 5 = 0
Solution
Given equation, 2x2 + 7x + 5 = 0
Here, a = 2, b = 7 and c = 5
By quadratic formula, we have
∴ x = -1 or x = -5/2
(vii) 2/3 x = -1/6 x2 – 1/3
Solution
Given equation, 2/3 x = -1/6 x2 – 1/3
1/6 x2 + 2/3 x + 1/3 = 0
Multiplying by 6 on both sides
x2 + 4x + 2 = 0
Here, a = 1, b = 4 and c = 2
By quadratic formula, we have
∴ x = -2 + √2 or x = -2 – √2
(viii) 1/15 x2 + 5/3 = 2/3 x
Solution
Given equation, 1/15 x2 + 5/3 = 2/3 x
1/15 x2 – 2/3 x + 5/3 = 0
Multiplying by 15 on both sides
x2 – 10x + 25 = 0
Here, a = 1, b = -10 and c = 25
By quadratic formula, we have
∴ x = 5 (equal roots)
(ix) x2 – 6 = 2 √2 x
Solution
Given equation, x2 – 6 = 2 √2 x
x2 – 2√2 x – 6 = 0
Here, a = 1, b = -2√2 and c = -6
By quadratic formula, we have
∴ x = 3√2 or x = -√2
(x) 4/x – 3 = 5/ (2x + 3)
Solution
Given equation, 4/x – 3 = 5/ (2x + 3)
(4 – 3x)/ x = 5/ (2x + 3)
On cross multiplying, we have
(4 – 3x)(2x + 3) = 5x
⇒ 8x – 6x2 + 12 – 9x = 5x
⇒ 6x2 + 6x – 12 = 0
Dividing by 6, we get
x2 + x – 2 = 0
Here, a = 1, b = 1 and c = -2
By quadratic formula, we have
∴ x = 1 or x = -2
(xi) 2x + 3/ x + 3 = x + 4/ x + 2
Solution
Given equation, 2x + 3/ x + 3 = x + 4/ x + 2
On cross-multiplying, we have
(2x + 3) (x + 2) = (x + 4) (x + 3)
⇒ 2x2 + 4x + 3x + 6 = x2 + 3x + 4x + 12
⇒ 2x2 + 7x + 6 = x2 + 7x + 12
⇒ x2 + 0x – 6 = 0
Here, a = 1, b = 0 and c = -6
By quadratic formula, we have
∴ x = √6 or x = -√6
(xii) √6x2 – 4x – 2√6 = 0
Solution
Given equation, √6x2 – 4x – 2√6 = 0
Here, a = √6, b = -4 and c = -2√6
By quadratic formula, we have
∴ x = √6 or -√6/3
(xiii) 2x/ x – 4 + (2x – 5)/(x – 3) =
Solution
Given equation, 2x/ x – 4 + (2x – 5)/(x – 3) =
⇒ 25x2 – 175x + 300 = 12x2 – 57x + 60
⇒ 13x2 – 118x + 240 = 0
Here, a = 13, b = -118 and c = 240
By quadratic formula, we have
∴ x = 6 or x = 40/13
(xiv) (x-1)/(x-2) + (x-3)/(x-4) = 3(1/3)
Solution
From the given equation,
⇒ 10x2 – 60x + 80 = 6x2 – 30x + 30
⇒ 4x2 – 30x + 50 = 0
⇒ 2x2 – 15x + 25 = 0
Here, a = 2, b = -15 and c = 25
∴ x = 5 or x = 5/2
2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:
(i) x2 – 8x +5 = 0
(ii) 5x2 + 10x – 3 = 0
Solution
(i) x2 – 8x + 5 = 0
Here, a = 1, b = -8 and c = 5
By quadratic formula, we have
x = 4 ± 3.3
Thus, x = 7.7 or x = 0.7
(ii) 5x2 + 10x – 3 = 0
Here, a = 5, b = 10 and c = -3
By quadratic formula, we have
Thus, x = 0.3 or x = -2.3
3. Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places:
(i) 2x2 – 10x + 5 = 0
Solution
Given equation, 2x2 – 10x + 5 = 0
Here, a = 2, b = -10 and c = 5
∴ x = 4.44 or x = 0.56
(ii) 4x + 6/x + 13 = 0
Solution
Given equation, 4x + 6/x + 13 = 0
Multiplying by x both sides, we get
4x2 + 13x + 6 = 0
Here, a = 4, b = 13 and c = 6
∴ x = -0.56 or x = -2.70
(iii) 4x2 – 5x – 3 = 0
Solution
Given equation, 4x2 – 5x – 3 = 0
Here, a = 4, b = -5 and c = -3
∴ x = 1.70 or x = -0.44
(iv) x2 – 3x – 9 = 0
Solution
Given equation, x2 – 3x – 9 = 0
Here, a = 1, b = -3 and c = -9
∴ x = 4.85 or x = -1.85
(v) x2 – 5x – 10 = 0
Solution
Given equation, x2 – 5x – 10 = 0
Here, a = 1, b = -5 and c = -10
∴ x = 6.53 or x = -1.53
4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:
(i) 3x2 – 12x – 1 = 0
(ii) x2 – 16 x +6 = 0
(iii) 2x2 + 11x + 4 = 0
Solution
(i) Given equation, 3x2 – 12x – 1 = 0
Here, a = 3, b = -12 and c = -1
∴ x = 4.082 or x = -0.082
(ii) Given equation, x2 – 16 x + 6 = 0
Here, a = 1, b = -16 and c = 6
∴ x = 15.616 or x = 0.384
(iii) Given equation, 2x2 + 11x + 4 = 0
Here, a = 2, b = 11 and c = 4
∴ x = -0.392 or x = -5.110
5. Solve:
(i) x4 – 2x2 – 3 = 0
Solution
Given equation, x4 – 2x2 – 3 = 0
⇒ x4 – 3x2 + x2 – 3 = 0
⇒ x2(x2 – 3) + 1(x2 – 3) = 0
⇒ (x2 + 1) (x2 – 3) = 0
So, x2 + 1 = 0 (which is not possible) or x2 – 3 = 0
Hence, x2 – 3 = 0
⇒ x = ± √3
(ii) x4 – 10x2 + 9 = 0
Solution
Given equation, x4 – 10x2 + 9 = 0
⇒ x4 – x2 – 9x2+ 9 = 0
⇒ x2(x2 – 1) – 9(x2 – 1) = 0
⇒ (x2 – 9)(x2 – 1) = 0
So, we have
x2 – 9 = 0 or x2 – 1 = 0
Hence, x = ± 3 or x = ± 1
Exercise 5(E)
1. Solve each of the following equations:
2x/(x-3) + 1/(2x+3) + (3x+9)/(x-3)(2x+3) = 0; x ≠ 3, x ≠ -3/2
Solution
Given equation,
⇒ 4x2 + 6x + x – 3 + 3x + 9 = 0
⇒ 4x2 + 10x + 6 = 0
⇒ 4x2 + 4x + 6x + 6 = 0
⇒ 4x(x + 1) + 6(x + 1) = 0
⇒ (4x + 6) (x + 1) = 0
So, 4x + 6 = 0 or x + 1 = 0
x = -1 or x = -6/4 = -3/2 (rejected as this value is excluded in the domain)
∴ x = -1 is the only solution
2. (2x + 3)2 = 81
Solution
Given, (2x + 3)2 = 81
Taking square root on both sides we have,
2x + 3 = ± 9
⇒ 2x = ± 9 – 3
⇒ x = (± 9 – 3)/ 2
So, x = (9 – 3)/ 2 or (-9 – 3)/ 2
∴ x = 3 or x = -6
3. a2x2 – b2 = 0
Solution
Given equation, a2x2 – b2 = 0
⇒ (ax)2 – b2 = 0
⇒ (ax + b)(ax – b) = 0
So, ax + b = 0 or ax – b = 0
∴ x = -b/a or b/a
4. x2 – 11/4 x + 15/8 = 0
Solution
Given equation, x2 – 11/4 x + 15/8 = 0
Taking L.C.M we have,
(8x2 – 22x + 15)/ 8 = 0
⇒ 8x2 – 22x + 15 = 0
⇒ 8x2 – 12x – 10x + 15 = 0
⇒ 4x(2x – 3) – 5(2x – 3) = 0
⇒ (4x – 5)(2x – 3) = 0
So, 4x – 5 = 0 or 2x – 3 = 0
∴ x = 5/4 or x = 3/2
5. x + 4/x = -4; x ≠ 0
Solution
Given equation, x + 4/x = -4
⇒ (x2 + 4)/ x = -4
⇒ x2 + 4 = -4x
⇒ x2 + 4x + 4 = 0
⇒ x2 + 2x + 2x + 4 = 0
⇒ x(x + 2) + 2(x + 2) = 0
⇒ (x + 2)(x + 2) = 0
⇒ (x + 2)2 = 0
Taking square–root we have,
x + 2 = 0
∴ x = -2
6. 2x4 – 5x2 + 3 = 0
Solution
Given equation, 2x4 – 5x2 + 3 = 0
Let’s take x2 = y
Then, the equation becomes
2y2 – 5y + 3 = 0
⇒ 2y2 – 2y – 3y + 3 = 0
⇒ 2y(y – 1) – 3(y – 1) = 0
⇒ (2y – 3) (y – 1) = 0
So, 2y – 3 = 0 or y – 1 = 0
y = 3/2 or y = 1
And, we have taken y = x2
Thus,
x2 = 3/2 or x2 = 1
⇒ x = ± √(3/2) or x = ±1
7. x4 – 2x2 – 3 = 0
Solution
Given equation, x4 – 2x2 – 3 = 0
Let’s take x2 = y
Then, the equation becomes
y2 – 2y – 3 = 0
⇒ y2 – 3y + y – 3 = 0
⇒ y(y – 3) + 1(y – 3) = 0
⇒ (y + 1)(y – 3) = 0
So, y + 1 = 0 or y – 3 = 0
⇒ y = -1 or y = 3
And, we have taken y = x2
Thus, x2 = – 1(impossible, no real solution)
⇒ x2 = 3
⇒ x = ± √3
8. 9(x2 + 1/x2)− 9(x+1/x) – 52 = 0
Solution
Let us take (x + 1/x) = y …. (1)
Now, squaring it on both sides
(x + 1/x)2 = y2
⇒ x2 + 1/x2 + 2 = y2
So, x2 + 1/x2 = y2 – 2 ... (2)
Using (1) and (2) in the given equation, we have
9(y2 – 2) – 9(y) – 52 = 0
⇒ 9y2 – 18 – 9y – 52 = 0
⇒ 9y2 – 9y – 70 = 0
⇒ 9y2 – 30y + 21y – 70 = 0
⇒ 3y(3y – 10) + 7(3y – 10) = 0
⇒ (3y + 7)(3y – 10) = 0
So, 3y + 7 = 0 or 3y – 10 = 0
⇒ y = -7/3 or y = 10/3
Now,
x + 1/x = -7/3 or x + 1/x = 10/3
⇒ (x2 + 1)/x = -7/3 or (x2 + 1)/x = 10/3
⇒ 3x2 – 10x + 3 = 0 or 3x2 + 7x + 3 = 0
⇒ 3x2 – 9x – x + 3 = 0 or
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1)(x – 3) = 0
So, x = 1/3 or 3
9. 2(x2 + 1/x2)− (x+1/x) = 11
Solution
Let us take (x + 1/x) = y …. (1)
Now, squaring it on both sides
(x + 1/x)2 = y2
⇒ x2 + 1/x2 + 2 = y2
So,
x2 + 1/x2 = y2 – 2 ….. (2)
Using (1) and (2) in the given equation, we have
2(y2 – 2) – (y) = 11
⇒ 2y2 – 4 – y = 11
⇒ 2y2 – y – 15 = 0
⇒ 2y2 – 6y + 5y – 15 = 0
⇒ 2y(y – 3) + 5(y – 3) = 0
⇒ (2y + 5) (y – 3) = 0
So, 2y + 5 = 0 or y – 3 = 0
⇒ y = -5/2 or y = 3
Now, x + 1/x = -5/2 or x + 1/x = 3
⇒ (x2 + 1)/x = -5/2 or (x2 + 1)/x = 3
⇒ 2(x2 + 1) = -5x or x2 + 1 = 3x
2x2 + 5x + 2 = 0 or x2 – 3x + 1 = 0
⇒ 2x2 + 4x + x + 2 = 0 or
⇒ 2x(x + 2) + 1(x + 2) = 0
⇒ (2x + 1)(x + 2) = 0
Hence, x = -1/2 or -2
10. (x2 + 1/x2)− 3(x−1/x) −2 = 0
Solution
Let us take (x – 1/x) = y …. (1)
Now, squaring it on both sides
(x – 1/x)2 = y2
⇒ x2 + 1/x2 – 2 = y2
So, x2 + 1/x2 = y2 + 2 ….. (2)
Using (1) and (2) in the given equation, we have
(y2 + 2) – 3(y) – 2 = 0
⇒ y2 -3y = 0
⇒ y(y – 3) = 0
So, y = 0 or y – 3 = 0
Now,
(x – 1/x) = 0 or (x – 1/x) = 3
⇒ x2 – 1 = 0 or x2 – 1 = 3x
⇒ x2 = 1 or x2 – 3x – 1 = 0
∴ x = ± 1
Exercise 5(F)
1. Solve:
(i) (x + 5) (x – 5) = 24
Solution
Given equation, (x + 5) (x – 5) = 24
⇒ x2 – 25 = 24
⇒ x2 = 49
Thus, x = ± 7
(ii) 3x2 – 2√6x + 2 = 0
Solution
Given equation, 3x2 – 2√6x + 2 = 0
⇒ 3x2 – √6x – √6x + 2 = 0
⇒ √3x(√3x – √2) – √2(√3x – √2) = 0
⇒ (√3x – √2) (√3x – √2) = 0
So, √3x – √2 = 0 or √3x – √2 = 0
∴ x = √(2/3), √(2/3) (equal roots)
(iii) 3√2x2 – 5x – √2 = 0
Solution
Given equation, 3√2x2 – 5x – √2 = 0
⇒ 3√2x2 – 6x + x – √2 = 0
⇒ 3√2x(x – √2) + 1(x – √2) = 0
⇒ (3√2x + 1) (x – √2) = 0
So, 3√2x + 1 = 0 or x – √2 = 0
∴ x = -1/ 3√2 or x = √2
(iv) 2x – 3 = √(2x2 – 2x + 21)
Solution
Given equation, 2x – 3 = √(2x2 – 2x + 21)
On squaring on both sides, we have
(2x – 3)2 = 2x2 – 2x + 21
⇒ 4x2 + 9 – 12x = 2x2 – 2x + 21
⇒ 2x2 – 10x – 12 = 0
Dividing by 2, we get
x2 – 5x – 6 = 0
⇒ x2 – 6x + x – 6 = 0
⇒ x(x – 6) + 1(x – 6) = 0
⇒ (x + 1) (x – 6) = 0
So, x + 1 = 0 or x – 6 = 0
Thus, we get
x = -1 or x = 6
But, putting x = -1 the L.H.S become negative. And we know that the square root function always gives a positive value.
∴ x = 6 is the only solution.
2. One root of the quadratic equation 8x2 + mx + 15 = 0 is ¾. Find the value of m. Also, find the other root of the equation.
Solution
Given equation, 8x2 + mx + 15 = 0
One of the roots is ¾, and hence it satisfies the given equation
So,
8(3/4)2 + m(3/4) + 15 = 0
⇒ 8(9/16) + m(3/4) + 15 = 0
⇒ 18/4 + 3m/4 + 15 = 0
Taking L.C.M, we have
(18 + 3m + 60)/4 = 0
⇒ 18 + 3m + 60 = 0
⇒ 3m = – 78
⇒ m = -26
Now, putting the value of m in the given equation, we get
8x2 + (-26)x + 15 = 0
⇒ 8x2 – 26x + 15 = 0
⇒ 8x2 – 20x – 6x + 15 = 0
⇒ 4x(2x – 5) – 3(2x – 5) = 0
⇒ (4x – 3) (2x – 5) = 0
So, 4x – 3 = 0 or 2x – 5 = 0
∴ x = ¾ or x = 5/2
3. Show that one root of the quadratic equation x2 + (3 – 2a)x – 6a = 0 is -3. Hence, find its other root.
Solution
Given quadratic equation, x2 + (3 – 2a)x – 6a = 0
Now, putting x = -3 we have
(-3)2 + (3 – 2a)( -3) – 6a = 0
⇒ 9 – 9 + 6a – 6a = 0
⇒ 0 = 0
Since, x = -3 satisfies the given equation -3 is one of the root of the quadratic equation.
x2 + (3 – 2a)x – 6a = 0
⇒ x2 + 3x – 2ax – 6a = 0
⇒ x(x + 3) – 2a(x + 3) = 0
⇒ (x – 2a) (x + 3) = 0
So, x – 2a = 0 or x + 3 =0
x = 2a or x = -3
Hence, the other root is 2a.
4. If p – 15 = 0 and 2x2 + px + 25 = 0: find the values of x.
Solution
Given equations, p – 15 = 0 and 2x2 + px + 25 = 0
Thus, p = 15
Now, using p in the quadratic equation, we get
2x2 + (15)x + 25 = 0
⇒ 2x2 + 10x + 5x + 25 = 0
⇒ 2x(x + 5) + 5(x + 5) = 0
⇒ (2x + 5) (x + 5) = 0
So, 2x + 5 = 0 or x + 5 = 0
Hence, x = -5/2 or x = -5
5. Find the solution of the quadratic equation 2x2 – mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.
Solution
Given,
m + 5 = 0 and n – 1 = 0
So, m = -5 and n = 1
Now, putting these values in the given quadratic equation 2x2 – mx – 25n = 0, we get
2x2 – (-5)x – 25(1) = 0
⇒ 2x2 + 5x – 25 = 0
⇒ 2x2 + 10x – 5x – 25 = 0
⇒ 2x(x + 5) -5(x + 5) = 0
⇒ (2x – 5) (x + 5) = 0
So, 2x – 5 = 0 or x + 5 = 0
Hence, x = 5/2 or x = -5
6. If m and n are roots of the equation: 1/x – 1/(x-2) = 3: where x ≠ 0 and x ≠ 2; find m x n.
Solution
Given equation, 1/x – 1/(x-2) = 3
⇒ (x – 2 – x)/ (x(x – 2)) = 3
⇒ -2 = 3(x2 – 2x)
⇒ 3x2 – 6x + 2 = 0
Solving by using quadratic formula, we get
And, since m and n are roots of the equation, we have
m = (√3 + 1)/ √3 n = (√3 – 1)/ √3
So, m×n = (√3 + 1)/ √3 × (√3 – 1)/ √3 = [(√3)2 – 1]/ (√3)2
Thus, m × n = 2/3